Answer:
V O2 = 1.623 L
Explanation:
- 1 mol ≡ 6.022 E23 molecules
∴ molecules O2 = 4.00 E22 molecules
⇒ moles O2 = (4.00 E22 molecules O2)×(mol O2/6.022 E23 molecules)
⇒ moles O2 = 0.0664 moles
at STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
assuming ideal gas:
∴ V = RTn/P
⇒ V O2 = ((0.082 atm.L/K.mol)(298 K)(0.0664 mol))/( 1 atm)
⇒ V O2 = 1.623 L
Regard the principle of utilization of two gas.
Make a consistent control of hardware containing gas.
Make a consistent control of weight diminishing valves giving gas.
No smoking zone.
Answer:
= 13.0 moles O2
Explanation:
1] Given the equation: 2C8H18 + 25 O2 ----> 16CO2 + 18H2O
a. How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide?
8.33 moles CO2 X
25mol O2
16mol CO2
= 13.0 moles O2
It would contain 0.105M moles of NaOH in each liter of solution
